anonymous
  • anonymous
Hi all, I am still working on second derivative tests having trouble with finding the critical points. The problem is find the critical points of f(x,y)=sinx+siny+cos(x+y) for x between 0 and pie/4 and y between 0 and pie/4. Classify each as a local min, max, or saddle point. f(x,y)=sinx+siny+cos(x+y) I found the partials to be: derivative of f with respect to x = fx=cosx-sin(x+y) derivative of f with respect to y = fy=cosy-sin(x+y) I am having trouble solving the partials for 0.
MIT 18.02 Multivariable Calculus, Fall 2007
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
For fx=cosx-sin(x+y)=0 fy=cosy-sin(x+y)=0 x=pie/6 y=pie/6 I did this by trial and error. Does anyone know how to solve it algebraically?
anonymous
  • anonymous
First, eliminate sin(x+y): fx-fy=0 cos(x)-cos(y)=0 cos(x)=cos(y) x=y Plug x=y to either fx or fy: fx=cos(y)-sin(y+y)=0 cos(y)=sin(2y) cos(y)=cos(pi/2-2y) y=(pi/2)-2y 3y=pi/2 y=pi/6 Since x=y, x=pi/6
anonymous
  • anonymous
Very clever!! That makes a lot of sense!! It's amazing how easy it looks once you've seen it done. I have been puzzling over that one for days! Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Ur welcome! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.